The three vertices of an equilateral triangle are at distances 1, 2, and 3 from a line.
Find all possible areas of this triangle.
I did try to get a generalized solution for three arbitrary lines, but that got messy. However the specific case of all three distances on the same side of the line (which I will call the directrix) had a very nice elementary geometric solution.
Let A be the vertex with distance 1 from the directrix. Similarly, let B and C be the vertices with distance 2 and 3 respectively.
Now let point M be the midpoint of AC. Then M is distance 2 from the directrix. Draw median BM. Since both endpoints are the same distance from the directrix then BM is parallel to the directrix.
Since ABC is equilateral median BM is also an altitude. Then AC is perpendicular to the directrix. Then this means the length of AC is simply 3-1=2. The area of an equilateral triangle 2 is (sqrt(3)/4)*x^2, and in this case x=2 so the area is sqrt(3).
Thoughts
